A body is loaded elastically so that strain in y-direction (e_y) = 0 and σ_z = 0. Take σ_x and σ_y as finite. What is the value of strain in z-direction (e_z) in terms of stress in x-direction (σ_x), Young's Modulus (E), and Poisson's ratio (v)?

Explanation:

Concept:

\(e_v=e_x+e_y+e_z\)

\(e_x=\frac{\sigma_x}{E}-\frac{μ\sigma_y}{E}-\frac{μ\sigma_z}{E}\)

\(e_y=\frac{\sigma_y}{E}-\frac{μ\sigma_x}{E}-\frac{μ\sigma_z}{E}\)

\(e_y=\frac{\sigma_x}{E}-\frac{μ\sigma_x}{E}-\frac{μ\sigma_y}{E}\)

Calculation:

As we know,

Strain in y-direction \(e_y=\frac{\sigma_y}{E}-\frac{μ\sigma_x}{E}-\frac{μ\sigma_z}{E}\)

where σ_x, σ_y and σ_z are normal stresses in x, y and z-directions.

So, \(0=\frac{\sigma_y}{E}-\frac{μ\sigma_x}{E}\)

\(\frac{\sigma_y}{E}=\frac{μ\sigma_x}{E}\)

Now, strain in z-direction, 

\(e_z=\frac{\sigma_z}{E}-\frac{μ\sigma_x}{E}-\frac{μ\sigma_y}{E}\)

using above relation         

\(e_z=0-μ(\frac{\sigma_x}{E}+\frac{μ\sigma_x}{E})\)

\(e_z=-μ\sigma_x(\frac{1+μ}{E})\) or ez = -σxv(1 + v)/E { where μ = v}